1.) A biochemistry student wants to make 4 liters of a phosphate solution using 48.6 g Na2HPO4 (FW=141.96) What is the concentration of this solution?
2. When she reaches the storeroom she finds that there is no Na2HPO4, however there is a bottle of hydrate Na2HPO4 * 7H20. (FW=268.07) how many grams of this salt should she use in order to make 4 liters with the same phosphate concentration as above?
1)
Given:
V = 4[L]
m1 = 48.6[g]
w1 = 141.96[g/mol]
Use:
c = m1/(w1*V)
2)
Given:
w2 = 268.07
Use:
m2/(w2*V) = m1/(w1*V)
.: m2 = m1*w2/w1
I'm actually still confused by this
To find the concentration of a solution, we need to determine the number of moles of the solute (in this case, Na2HPO4) and the volume of the solution.
1. Concentration of the Na2HPO4 solution using 48.6 g:
- First, we need to convert the mass of Na2HPO4 to moles by dividing it by its molar mass.
Mass = 48.6 g
Molar mass of Na2HPO4 = 141.96 g/mol
Moles = Mass / Molar mass = 48.6 g / 141.96 g/mol = 0.342 mol
- The volume of the solution is given as 4 liters.
- Concentration(C) = Moles / Volume
= 0.342 mol / 4 L = 0.0855 mol/L
Therefore, the concentration of the Na2HPO4 solution is 0.0855 mol/L.
2. Now, let's find out the amount of hydrate Na2HPO4 * 7H2O required to make a 4L solution.
- The molecular weight of the hydrate Na2HPO4 * 7H2O is given as 268.07 g/mol.
- We know that the concentration of the solution should remain the same as in question 1.
- Concentration = 0.0855 mol/L
- Volume = 4 L
- Moles = Concentration * Volume
= 0.0855 mol/L * 4 L = 0.342 mol
- The molar ratio between anhydrous Na2HPO4 and hydrate Na2HPO4 * 7H2O is 1:1 (one mole to one mole).
- Therefore, to obtain the same number of moles, she should use 0.342 mol of the hydrate form of Na2HPO4 * 7H2O.
- Now, to find the mass, we will multiply the moles by the molar mass.
- Mass = Moles * Molar mass = 0.342 mol * 268.07 g/mol = 91.81 g
Therefore, she should use 91.81 g of Na2HPO4 * 7H2O to make a 4-liter solution with the same phosphate concentration as above.